PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Ta có: \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2\left(LT\right)}=\dfrac{1}{2}n_{KMnO_4}=0,1\left(mol\right)\)
Mà: H% = 80%
\(\Rightarrow n_{O_2\left(TT\right)}=0,1.80\%=0,08\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,08.22,4=1,792\left(l\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{4}>\dfrac{0,08}{5}\), ta được P dư.
Theo PT: \(n_{P\left(pư\right)}=\dfrac{4}{5}n_{O_2}=0,064\left(mol\right)\)
\(\Rightarrow n_{P\left(dư\right)}=0,1-0,064=0,036\left(mol\right)\)
\(\Rightarrow m_{P\left(dư\right)}=0,036.31=1,116\left(g\right)\)
Bạn tham khảo nhé!